... by: Jools
Further to Sean's post, I've also found [at least] two examples of such a "diagonal" hidden non-unique rectangle that could be added to your solver. Both examples satisfy the existing NUR rules. You should be able to step through both grids in the solver until coming across an XY-wing, when they should become visible.
Type (i) 5/8 in F1 and G3; entering a 5 in F3 would force G1 to become one too - thus forming a "deadly" pattern, so the 5 in F3 can be removed.
Doing this enabled me to breeze through the rest of the puzzle (after also completing the XY-wing on C2-H2-G3 which removed the 5s in B3 and C3), because it immediately left just a single 5 in G3 for column 3.
Note that type (i) candidates do not have to form part of a N=2x2 pattern.
This one actually has two types at once; the second removes two candidates!
Type (i) 2/6 in D7 and F5; entering a 2 in F7 would force D5 to become one too - thus forming a "deadly" pattern, so the 2 in F7 can be removed.
Type (ii) 6/7 in G6 and H9; entering a 7 in either G9 or H6 would force the other to be the same - again forming a "deadly" pattern, so both 7s in G9 and H6 can be removed.
Note that type (ii) candidates *do* form a N=2x2 pattern, which enables both corners to have a candidate removed.
Both examples come from my Nintendo DS Platinum Sudoku cartridge, from the 5,000,000 grids on "Pro" level. (I'm up to #3293 so far!)
... by: Sean Tremba
I think I've come up with a variant on unique rectangles that doesn't seem to be covered by any of your HUR or UR strategy types. (You can certainly correct me if I'm wrong).
Suppose you have an "x-wing" formation on a given candidate, and you have identical bi-value cells diagonally opposed to each other in that formation. That is, you have two cells with xy as candidates at diagonal corners. The other two cells then must contain x with some other set of candidates (which may or may not include y). I maintain that you can then conclude that x is the solution to the two diagonal cells which have xy as candidates.
To prove this, assume that x is not the solution to these diagonally opposed cells. x is therefore solution to the other two cells in the "x wing" formation. Since the only other candidate for these cells is y, y must then be their solution. However, this leads to a deadly pattern, since you have a rectangle with y as the solution to the cells on one diagonal and x as the solution on the other; x and y could be reversed to give a second valid solution.
More concretely, (and also the example I found), suppose cells F1 and H2 are bivalue cells with 7 and 9 as candidates. In the sudoku I was working, cell F2 had candidates 1,7 and 9, and cell H1 had candidates 7,8 and 9. These cells were the only cells on rows F and H which contained 7's, and the only cells in columns 1 and 2 which contained 7's. The solution to cells F1 and H2 must then be 7. Otherwise, the solution to F1 and H2 would be 9, and the solution to F2 and H1 would be 7, which creates a deadly pattern.
... by: Chuck Bruno - Virginia
After importing the puzzle below and enabling only X-Wing, Y-Wing, Sword-Fish, Unique Rectangles, Hidden Unique Rectangles, and XYZ Wing/WXYZ Wing, stepping through the puzzle eventually yields a Type 1 Hidden Unique Rectangle at AC15 which enables the removal of 7 at C1. I can't see it, can you please help?